Can We Cancel Those Zeroes?

Geometry Level 3

sin ⁡ 2 0 ∘ sin ⁡ 4 0 ∘ sin ⁡ 6 0 ∘ sin ⁡ 8 0 ∘ sin ⁡ 1 0 ∘ sin ⁡ 3 0 ∘ sin ⁡ 5 0 ∘ sin ⁡ 7 0 ∘ = ? \large \dfrac{\sin 20^\circ \sin 40^\circ \sin 60 ^\circ\sin 80^\circ}{\sin 10^\circ \sin 30^\circ \sin 50 ^\circ\sin 70^\circ} = \, ?


The answer is 3.

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Rohit Udaiwal by Rohit Udaiwal , 20, India

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1 solution

X = sin ⁡ 2 0 ∘ sin ⁡ 4 0 ∘ sin ⁡ 6 0 ∘ sin ⁡ 8 0 ∘ sin ⁡ 1 0 ∘ sin ⁡ 3 0 ∘ sin ⁡ 5 0 ∘ sin ⁡ 7 0 ∘ Note that: sin ⁡ ( θ ) = cos ⁡ ( 9 0 ∘ − θ ) = sin ⁡ 2 0 ∘ sin ⁡ 4 0 ∘ sin ⁡ 6 0 ∘ sin ⁡ 8 0 ∘ cos ⁡ 8 0 ∘ cos ⁡ 6 0 ∘ cos ⁡ 4 0 ∘ cos ⁡ 2 0 ∘ = tan ⁡ 2 0 ∘ tan ⁡ 4 0 ∘ tan ⁡ 6 0 ∘ tan ⁡ 8 0 ∘ = tan ⁡ 6 0 ∘ tan ⁡ 2 0 ∘ tan ⁡ ( 6 0 ∘ − 2 0 ∘ ) tan ⁡ ( 6 0 ∘ + 2 0 ∘ ) = tan ⁡ 6 0 ∘ tan ⁡ 6 0 ∘ Note that: tan ⁡ ( 3 θ ) = tan ⁡ θ tan ⁡ ( 6 0 ∘ − θ ) tan ⁡ ( 6 0 ∘ + θ ) = tan ⁡ 2 6 0 ∘ = 3 \begin{aligned} X & = \frac {\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ} {\color{#3D99F6}{\sin 10^\circ \sin 30^\circ \sin 50^\circ \sin 70^\circ}} \quad \quad \small \color{#3D99F6}{\text{Note that: } \sin(\theta) = \cos\left(90^\circ-\theta \right)} \\ & = \frac {\sin 20^\circ \sin 40^\circ \sin 60^\circ \sin 80^\circ} {\color{#3D99F6}{\cos 80^\circ \cos 60^\circ \cos 40^\circ \cos 20^\circ}} \\ & = \color{#3D99F6}{\tan 20^\circ \tan 40^\circ} \tan 60^\circ \color{#3D99F6}{\tan 80^\circ} \\ & = \tan 60^\circ \color{#3D99F6}{\tan 20^\circ \tan (60^\circ - 20^\circ) \tan (60^\circ + 20^\circ)} \\ & = \tan 60^\circ \color{#3D99F6}{\tan 60^\circ} \quad \quad \small \color{#3D99F6}{\text{Note that: } \tan (3 \theta) = \tan \theta \tan (60^\circ - \theta) \tan (60^\circ + \theta)} \\ & = \tan^2 60^\circ = \boxed {3} \end{aligned}

9 Helpful 1 Interesting 0 Brilliant 0 Confused

Exactly , A simple one though. I'm an 8th grader and we completed Trigonometry yesterday.

Abhiram Rao - 5 years, 2 months ago

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